3.6.7 \(\int \frac {\sqrt {f-c f x} (a+b \text {ArcSin}(c x))}{\sqrt {d+c d x}} \, dx\) [507]

Optimal. Leaf size=141 \[ -\frac {b f x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}} \]

[Out]

f*(-c^2*x^2+1)*(a+b*arcsin(c*x))/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)-b*f*x*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(
-c*f*x+f)^(1/2)+1/2*f*(a+b*arcsin(c*x))^2*(-c^2*x^2+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*f*x+f)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4763, 4847, 4737, 4767, 8} \begin {gather*} \frac {f \sqrt {1-c^2 x^2} (a+b \text {ArcSin}(c x))^2}{2 b c \sqrt {c d x+d} \sqrt {f-c f x}}+\frac {f \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))}{c \sqrt {c d x+d} \sqrt {f-c f x}}-\frac {b f x \sqrt {1-c^2 x^2}}{\sqrt {c d x+d} \sqrt {f-c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/Sqrt[d + c*d*x],x]

[Out]

-((b*f*x*Sqrt[1 - c^2*x^2])/(Sqrt[d + c*d*x]*Sqrt[f - c*f*x])) + (f*(1 - c^2*x^2)*(a + b*ArcSin[c*x]))/(c*Sqrt
[d + c*d*x]*Sqrt[f - c*f*x]) + (f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(2*b*c*Sqrt[d + c*d*x]*Sqrt[f - c*f
*x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4737

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*Si
mp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c
^2*d + e, 0] && NeQ[n, -1]

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4767

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x^2)^(
p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Dist[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p
], Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*
d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

Rule 4847

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rubi steps

\begin {align*} \int \frac {\sqrt {f-c f x} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d+c d x}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {(f-c f x) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {\sqrt {1-c^2 x^2} \int \left (\frac {f \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}-\frac {c f x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}}\right ) \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {\left (f \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}-\frac {\left (c f \sqrt {1-c^2 x^2}\right ) \int \frac {x \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=\frac {f \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}}-\frac {\left (b f \sqrt {1-c^2 x^2}\right ) \int 1 \, dx}{\sqrt {d+c d x} \sqrt {f-c f x}}\\ &=-\frac {b f x \sqrt {1-c^2 x^2}}{\sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt {d+c d x} \sqrt {f-c f x}}+\frac {f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{2 b c \sqrt {d+c d x} \sqrt {f-c f x}}\\ \end {align*}

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Mathematica [A]
time = 0.39, size = 200, normalized size = 1.42 \begin {gather*} \frac {\frac {2 \sqrt {d+c d x} \sqrt {f-c f x} \left (-b c x+a \sqrt {1-c^2 x^2}\right )}{\sqrt {1-c^2 x^2}}+2 b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)+\frac {b \sqrt {d+c d x} \sqrt {f-c f x} \text {ArcSin}(c x)^2}{\sqrt {1-c^2 x^2}}-2 a \sqrt {d} \sqrt {f} \text {ArcTan}\left (\frac {c x \sqrt {d+c d x} \sqrt {f-c f x}}{\sqrt {d} \sqrt {f} \left (-1+c^2 x^2\right )}\right )}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[f - c*f*x]*(a + b*ArcSin[c*x]))/Sqrt[d + c*d*x],x]

[Out]

((2*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*(-(b*c*x) + a*Sqrt[1 - c^2*x^2]))/Sqrt[1 - c^2*x^2] + 2*b*Sqrt[d + c*d*x]*
Sqrt[f - c*f*x]*ArcSin[c*x] + (b*Sqrt[d + c*d*x]*Sqrt[f - c*f*x]*ArcSin[c*x]^2)/Sqrt[1 - c^2*x^2] - 2*a*Sqrt[d
]*Sqrt[f]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[f - c*f*x])/(Sqrt[d]*Sqrt[f]*(-1 + c^2*x^2))])/(2*c*d)

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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right ) \sqrt {-c f x +f}}{\sqrt {c d x +d}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))*(-c*f*x+f)^(1/2)/(c*d*x+d)^(1/2),x)

[Out]

int((a+b*arcsin(c*x))*(-c*f*x+f)^(1/2)/(c*d*x+d)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))*(-c*f*x+f)^(1/2)/(c*d*x+d)^(1/2),x, algorithm="maxima")

[Out]

a*(f*arcsin(c*x)/(c*d*sqrt(f/d)) + sqrt(-c^2*d*f*x^2 + d*f)/(c*d)) + b*sqrt(f)*integrate(sqrt(-c*x + 1)*arctan
2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/sqrt(c*x + 1), x)/sqrt(d)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))*(-c*f*x+f)^(1/2)/(c*d*x+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/sqrt(c*d*x + d), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- f \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}{\sqrt {d \left (c x + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))*(-c*f*x+f)**(1/2)/(c*d*x+d)**(1/2),x)

[Out]

Integral(sqrt(-f*(c*x - 1))*(a + b*asin(c*x))/sqrt(d*(c*x + 1)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))*(-c*f*x+f)^(1/2)/(c*d*x+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(-c*f*x + f)*(b*arcsin(c*x) + a)/sqrt(c*d*x + d), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x}}{\sqrt {d+c\,d\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(1/2),x)

[Out]

int(((a + b*asin(c*x))*(f - c*f*x)^(1/2))/(d + c*d*x)^(1/2), x)

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